The driver should be able to see the ground within ____

7. The "3 second rule" is the time you should pause at an intersection marked with a stop sign.

forsale [732]

A. True because that’s correct

8 0

3 years ago

P9.28 A large vacuum tank, held at 60 kPa absolute, sucks sea- level standard air through a converging nozzle whose throat diame

eimsori [14]

Answer:

a) $m=0.17kg/s$

b) $Ma=0.89$

Explanation:

From the question we are told that:

Pressure $P=60kPa$

Diameter $d=3cm$

Generally at sea level

$T_0=288k\\\\\rho_0=1.225kg/m^3\\\\P_0=101350Pa\\\\r=1.4$

Generally the Power series equation for Mach number is mathematically given by

$\frac{p_0}{p}=(1+\frac{r-1}{2}Ma^2)^{\frac{r}{r-1}}$

$\frac{101350}{60*10^3}=(1+\frac{1.4-1}{2}Ma^2)^{\frac{1.4}{1.4-1}}$

$Ma=0.89$

Therefore

Mass flow rate

$\frac{\rho_0}{\rho}=(1+\frac{1.4-1}{2}(0.89)^2)^{\frac{1.4}{1.4-1}}$

$\frac{1.225}{\rho}=(1+\frac{1.4-1}{2}(0.89)^2)^{\frac{1.4}{1.4-1}}$

$\rho=0.848kg/m^3$

Generally the equation for Velocity at throat is mathematically given by

$V=Ma(r*T_0\sqrt{T_e}$ )

Where

$T_e=\frac{P_e}{R\rho}\\\\T_e=\frac{60*10^6}{288*0.842\rho}$

$T_e=248$

Therefore

$V=0.89(1.4*288\sqrt{248})\\\\V=284$

Generally the equation for Mass flow rate is mathematically given by

$m=\rho*A*V$

$m=0.84*\frac{\pi}{4}*3*10^{-2}*284$

$m=0.17kg/s$

6 0

2 years ago

A cylindrical specimen of a hypothetical metal alloy is stressed in compression. If its original and final diameters are 30.00 a

IrinaVladis [17]

Answer:

The original length of the specimen $l_{o} = 104.7 mm$

Explanation:

Original diameter $d_{o}$ = 30 mm

Final diameter $d_{1}$ = 30.04 mm

Change in diameter Δd = 0.04 mm

Final length $l_{1}$ = 105.20 mm

Elastic modulus E = 65.5 G pa = 65.5 × $10^{3}$ M pa

Shear modulus G = 25.4 G pa = 25.4 × $10^{3}$ M pa

We know that the relation between the shear modulus & elastic modulus is given by

$G = \frac{E}{2(1 + \mu)}$

$25.5 = \frac{65.5}{2 (1 + \mu)}$

$\mu = 0.28$

This is the value of possion's ratio.

We know that the possion's ratio is given by

$\mu = \frac{\frac{0.04}{30} }{\frac{change \ in \ length}{l_{o} } }$

${\frac{change \ in \ length}{l_{o} } = \frac{\frac{0.04}{30} }{0.28}$

${\frac{change \ in \ length}{l_{o} } =$ 0.00476

$\frac{l_{1} - l_{o} }{l_{o} } = 0.00476$

$\frac{l_{1} }{l_{o} } = 1.00476$

Final length $l_{o}$ = 105.2 m

Original length

$l_{o} = \frac{105.2}{1.00476}$

$l_{o} = 104.7 mm$

This is the original length of the specimen.

5 0

3 years ago

The 40-ft-long A-36 steel rails on a train track are laid with a small gap between them to allow for thermal expansion. Determin

vfiekz [6]

Answer:

Ф = 0.02838 ft

F = 1,032 N

Explanation:

To find out gap delta,

As it is case of free thermal expansion,

First we start with, some assumptions we have to made to solve this problem.

1. Thermal Expansion Coefficient of Steel is ∝= 6.45 ×10^(-6)

2. Modulas of elasticity for A-36 steel is E= 200 GPa

3. Area of rail is assumed to be unit area.

The gape required can be given by,

Ф = ∝ × ΔT × L ... where Ф= Gap Delta in ft

ΔT= Temperature rise in F

= 90- (-20)

= 110 F

Ф = 6.45 ×10^(-6) × 110 × 40

Ф = 28,380 × 10^(-6) ft

Ф = 0.02838 ft .... total gape required for expansion of steel rails

Stress induced in rails is given by,

σ = ∝ × ΔT × E

= 6.45 ×10^(-6) × 110 × 200

σ = 1,41,900 Pa

Now, let's find axial force in rails,

Here,we have to consider ΔT= 20 F.

As due to temperature change, axial force generated in rails can be find by,

F = A × ∝ × ΔT× E × L

F = 1 × 6.45 × 10^(-6) × 20 × 200 × 10^(-9) × 40

F = 25,800 × 40 × 10^(-3)

F = 10,32,000 × 10^(-3)

F= 1,032 N

Finally, due to temperature change, rail is subjected to axial force, axial stress.

8 0

3 years ago

HELP PLEASE!!!!

DaniilM [7]

Answer:

DESCULPA MAS EU NÃO ENTENDI

8 0

3 years ago

The driver should be able to see the ground within _____ to the front?